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The restricted order of a basis is the least integer $t$ (if it exists) such that every large integer is the sum of at most $t$ distinct summands from $A$. What are necessary and sufficient conditions that this exists? Can it be bounded (when it exists) in terms of the order of the basis? What are necessary and sufficient conditions that this is equal to the order of the basis?
Bateman has observed that for $h\geq 3$ there is a basis of order $h$ with no restricted order, taking \[A=\{1\}\cup \{x>0 : h\mid x\}.\] Kelly [Ke57] has shown that any basis of order $2$ has restricted order at most $4$ and conjectures it always has restricted order at most $3$.

The set of squares has order $4$ and restricted order $5$ (see [Pa33]) and the set of triangular numbers has order $3$ and restricted order $3$ (see [Sc54]).

Is it true that if $A\backslash F$ is a basis for all finite sets $F$ then $A$ must have a restricted order? What if they are all bases of the same order?